Codeforces Round #442 (Div. 2) B. Nikita and string

http://codeforces.com/problemset/problem/877/B

One day Nikita found the string containing letters “a” and “b” only.

Nikita thinks that string is beautiful if it can be cut into 3 strings (possibly empty) without changing the order of the letters, where the 1-st and the 3-rd one contain only letters “a” and the 2-nd contains only letters “b”.

Nikita wants to make the string beautiful by removing some (possibly none) of its characters, but without changing their order. What is the maximum length of the string he can get?

Input

The first line contains a non-empty string of length not greater than 5 000 containing only lowercase English letters “a” and “b”.

Output

Print a single integer — the maximum possible size of beautiful string Nikita can get.

Examples

input

abba

output

4

input

bab

output

2

Note

It the first sample the string is already beautiful.

In the second sample he needs to delete one of “b” to make it beautiful.

题意

把一个字符串通过删去一些字符分成x*a+y*b+z*a的形式，问这个字符串最长几个字符

题解

用a[i]和b[i]记录i之前的a和b的数量，然后得到动态转移方程：
ans=max(ans,a[i]+(b[j]-b[i])+(a[l]-a[j]));

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;int main(){    char ch[5050];    scanf("%s",ch);    int l=strlen(ch);    int a[5050];    int b[5050];    a[0]=b[0]=0;    for(int i=0;i<l;i++)    {        a[i+1]=a[i]+(ch[i]=='a');        b[i+1]=b[i]+(ch[i]=='b');    }    int ans=0;    for(int i=0;i<=l;i++)        for(int j=i;j<=l;j++)            ans=max(ans,a[i]+(b[j]-b[i])+(a[l]-a[j]));        cout << ans << endl;        return 0;}

简单的方法

#include<bits/stdc++.h>using namespace std;int main(){    string s;    cin>>s;    int a=0,aba=0,ab=0;    int l=s.length();    for(int i=0; i<l; i++)    {        if(s[i]=='a') aba++,a++;        if(s[i]=='b') ab++;        aba=max(aba, ab);        ab=max(ab, a);    }    cout<<aba;}